Question: Find $AX$ in the diagram if $CX$ bisects $\angle ACB$.

[asy]
import markers;

real t=.56;
pair A=(0,0);
pair B=(3,2);
pair C=(.5,1.5);
pair X=t*A+(1-t)*B;

draw(C--A--B--C--X);

label("$A$",A,SW);
label("$B$",B,E);
label("$C$",C,N);
label("$X$",X,SE);

//markangle(n=1,radius=15,A,C,X,marker(markinterval(stickframe(n=1),true)));
//markangle(n=1,radius=15,X,C,B,marker(markinterval(stickframe(n=1),true)));

label("$28$",.5*(B+X),SE);
label("$30$",.5*(B+C),N);
label("$21$",.5*(A+C),NW);

[/asy]
Solution: The Angle Bisector Theorem tells us that  \[\frac{AC}{AX}=\frac{BC}{BX}\]so \[AX=\frac{AC\cdot BX}{BC}=\frac{21\cdot28}{30}=\frac{7^2\cdot3\cdot4}{30}=\frac{7^2\cdot2}{5}=\boxed{\frac{98}5}.\]